1909.]
The principle of relativity in electrodynamics.
97
areas, we have
l
d
s
L
d
S
=
r
2
R
2
=
1
Λ
2
,
{\displaystyle {\frac {l\ ds}{L\ dS}}={\frac {r^{2}}{R^{2}}}={\frac {1}{\Lambda ^{2}}},}
where
Λ
=
r
2
−
c
2
t
2
k
2
,
{\displaystyle \Lambda ={\frac {r^{2}-c^{2}t^{2}}{k^{2}}},}
n
d
s
N
d
S
=
m
d
s
M
d
S
=
r
δ
r
R
δ
R
,
{\displaystyle {\frac {n\ ds}{N\ dS}}={\frac {m\ ds}{M\ dS}}={\frac {r\ \delta r}{R\ \delta R}},}
in which
δ
r
,
δ
R
{\displaystyle \delta r,\delta R}
are corresponding elements of a radius,
δ
R
{\displaystyle \delta R}
being at rest in the (R, T ) system.
Hence
δ
r
δ
R
=
−
1
Λ
β
,
{\displaystyle {\frac {\delta r}{\delta R}}=-{\frac {1}{\Lambda \beta }},}
and
m
d
s
M
d
S
=
n
d
s
N
d
S
=
−
1
Λ
2
β
,
{\displaystyle {\frac {m\ ds}{M\ dS}}={\frac {n\ ds}{N\ dS}}=-{\frac {1}{\Lambda ^{2}\beta }},}
Using these and the final form of the transformation of the electric vectors in § 3, the following equations are obtained exactly as in the last section:
p
r
d
s
=
Λ
2
{
P
r
−
v
4
π
c
[
E
H
]
ν
}
d
S
,
p
θ
d
s
=
−
Λ
2
P
θ
d
S
/
β
,
p
ϕ
d
s
=
−
Λ
2
P
ϕ
d
S
/
β
,
{\displaystyle {\begin{array}{l}p_{r}ds=\Lambda ^{2}\left\{P_{r}-{\frac {v}{4\pi c}}[EH]_{\nu }\right\}dS,\\\\p_{\theta }ds=-\Lambda ^{2}P_{\theta }dS/\beta ,\\\\p_{\phi }ds=-\Lambda ^{2}P_{\phi }dS/\beta ,\end{array}}}
p and P being the mechanical forces per unit area according as the element is observed in the (r, t ) or (R, T ) system.
If the element dS belongs to the wall of a region in which there is radiation in equilibrium
[
E
H
]
ν
{\displaystyle [EH]_{\nu }}
is zero as before if the time-average be taken, and P is normal to dS .
Thus
p
r
d
s
=
Λ
2
P
L
d
S
=
Λ
4
P
l
d
s
,
p
θ
d
s
=
−
Λ
2
β
P
M
d
S
=
Λ
4
P
m
d
s
,
p
ϕ
d
s
=
−
Λ
2
β
P
N
d
S
=
Λ
4
P
n
d
s
.
{\displaystyle {\begin{array}{l}p_{r}ds=\Lambda ^{2}PL\ dS=\Lambda ^{4}Pl\ ds,\\\\p_{\theta }ds=-{\frac {\Lambda ^{2}}{\beta }}PM\ dS=\Lambda ^{4}Pm\ ds,\\\\p_{\phi }ds=-{\frac {\Lambda ^{2}}{\beta }}PN\ dS=\Lambda ^{4}Pn\ ds.\end{array}}}
Thus p is normal to ds and is equal to
Λ
4
P
.
{\displaystyle \Lambda ^{4}P.}
.
For the energy the transformation gives
ϵ
=
1
8
π
{
e
2
+
h
2
}
=
Λ
4
8
π
{
E
R
2
+
H
R
2
+
β
2
(
1
+
v
2
c
2
)
(
E
θ
2
+
E
ϕ
2
+
H
θ
2
+
H
ϕ
2
)
−
4
v
β
2
c
(
E
θ
H
ϕ
−
E
ϕ
H
θ
)
}
,
{\displaystyle {\begin{array}{ll}\epsilon &={\frac {1}{8\pi }}\left\{e^{2}+h^{2}\right\}\\\\&={\frac {\Lambda ^{4}}{8\pi }}\left\{E_{R}^{2}+H_{R}^{2}+\beta ^{2}\left(1+{\frac {v^{2}}{c^{2}}}\right)\left(E_{\theta }^{2}+E_{\phi }^{2}+H_{\theta }^{2}+H_{\phi }^{2}\right)-{\frac {4v\beta ^{2}}{c}}\left(E_{\theta }H_{\phi }-E_{\phi }H_{\theta }\right)\right\},\end{array}}}